What is the decimal for 2/3

The segment AC is?

Rama09 [41]

Answer:

B

Step-by-step explanation:

the hypotenuse........

6 0

2 years ago

Read 2 more answers

Suppose you work at a retail store that sells teddy bears. You always work 40 hours a week. The amount of your weekly paycheck i

brilliants [131]

a. Remember that f(x) is the amount of money you earned each week. Also note that teddybears = x

<em>f(x) = 5x + 300</em> is your equation

5x + 300 is your expression

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b. A reasonable domain can be 0 < x < 50, in which x is the amount you sell. 50 is a bit far out, but it takes into account holidays, as well as places you may sell at that have large amount of people.

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c. f(20) means that you sold 20 teddy bears, as 20 replaces x, which (like what was stated above) is the amount you sell

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d. Plug in 25 for x.

5x + 300 = t (total amount you make)

5(25) + 300 = t

Remember to follow PEMDAS. First, multiply 5 with 25

5 x 25 = 125

Next, add

125 + 300 = t

t = 425

$425 is the amount you make in a week.

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e. Note that you make $425 if you sell 25 teddy bears. Also remember that you are paid a flat $300. Remember the formula:

f(x) = 5x + 300

Plug in 425 for f(x)

425 = 5x + 300

Isolate the x. Do the opposite of PEMDAS. Note the equal sign. What you do to one side, you do to the other.

First, subtract 300 from both sides

425 (-300) = 5x + 300 (-300)

425 - 300 = 5x

125 = 5x

Next, divide 5 from both sides

(125)/5 = (5x)/5

x = 125/5

x = 25

25 teddy bears must be sold

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hope this helps

4 0

3 years ago

How do you solve <br> -7x+(-21)=41

alex41 [277]

Answer:

x = -62/7

Step-by-step explanation:

-7x+(-21)=41

-7x -21 =41

Add 21 to each side

-7x-21+21 = 41+21

-7x = 62

Divide by -7

-7x/-7 = 62/-7

x =- 62/7

7 0

4 years ago

Draw the graph for f(x)=x-1/x^2-x-6 and state:a. the x-intercept(s).b. the y-intercept.c. the equation(s) of any vertical asympt

Zigmanuir [339]

ANSWER:

a. (1, 0)

b. (0, 1/6)

c. x = -2. x = 3

d. y = 0

e. The asymptotes have the form of a line and divide the function into 3 parts.

f.

$\begin{equation*} \text{Domain: }\left(-\infty\:,\:-2\right)\cup\left(-2,\:3\right)\cup\left(3,\:\infty\:\right) \end{equation*}$

g.

$\text{Range}(-\infty,\infty)$

STEP-BY-STEP EXPLANATION:

We have the following function:

$f\left(x\right)=\frac{x-1}{x^2-x-6}$

The graph corresponding to the function is the following:

We determine in each case what the statement asks for, like this:

a. the x-intercept(s):

$\begin{gathered} \text{ in this case y = 0, therefore:} \\ \\ \frac{x-1}{x^2-x-6}=0 \\ \\ x-1=0\cdot(x^2-x-6) \\ \\ x=1 \\ \\ \text{The x-intercept is \lparen1, 0\rparen} \end{gathered}$

b. the y-intercept:

$\begin{gathered} \text{ in this case x = 0, therefore } \\ \\ y=\frac{0-1}{0^2-0-6} \\ \\ y=\frac{-1}{-6} \\ \\ y=\frac{1}{6} \\ \\ \text{ The y-intercept is }\:\left(0,\frac{1}{6}\right) \end{gathered}$

c. the equation(s) of any vertical asymptote(s). The horizontal asymptotes are the values that x cannot take since the function would be discontinuous for those values.

In this case, since it is a rational function, it would be when the denominator is 0, therefore, we solve the following:

$\begin{gathered} x^2-x-6=0 \\ \\ (x-3)(x+2)=0 \\ \\ x-3=0\rightarrow x=3 \\ \\ x+2=0\operatorname{\rightarrow}x=-2 \\ \\ \text{ The equation\lparen s\rparen of any vertical asymptote are:} \\ \\ x=3,x=-2 \end{gathered}$

d. the equation of the horizontal asymptote. If the degree of the denominator is greater than that of the numerator, the horizontal asymptote is the x-axis, that is:

$y=0$

e. information about the behavior at the asymptote(s).

In this case the behavior of the asymptotes are straight lines that represent values that the function cannot take or cannot reach, divide the function into 3 parts.

f and g.

In this case, the domain is the interval of values that x can take and the range is the interval of values that y can take.

Therefore:

$\begin{gathered} \text{Domain: }\left(-\infty\:,\:-2\right)\cup\left(-2,\:3\right)\cup\left(3,\:\infty\:\right) \\ \\ \text{Range:}\:\left(-\infty\:,\:\infty\:\right) \end{gathered}$